What is the change in volume for the reaction of $"16 g CH"_4$ and $"16 g O"_2$ at $700^@ "C"$ and $"1 bar"$ ? The reaction is $"CH"_4(g) + 1/2"O"_2(g) -> "CO"(g) + 2"H"_2(g)$ .

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What is the change in volume for the reaction of $"16 g CH"_4$ and $"16 g O"_2$ at $700^@ "C"$ and $"1 bar"$ ? The reaction is $"CH"_4(g) + 1/2"O"_2(g) -> "CO"(g) + 2"H"_2(g)$ .

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Answer 1 · 2017-04-11T15:30:54

About $"120 L"$ .

For this, since we are asked for the change in volume, it is necessary to solve for the molar volume of an ideal gas at $700^@ "C"$ (since it is evidently not $0^@ "C"$ like it would be for STP).

$PV = nRT$

$=> V/n = (RT)/P$

$= (("0.083145 L"cdot"bar/mol"cdot"K")("700+273.15 K"))/("1 bar")$

$=$ $"80.91 L/mol"$

Given $"16 g"$ of both methane and diatomic oxygen, we can then find their $"mol"$ s.

$"16 g CH"_4 xx ("1 mol CH"_4)/(12.011 + 4 xx 1.0079 "g CH"_4) = "0.9973 mols CH"_4$

$"16 g O"_2 xx ("1 mol O"_2)/(15.999 xx 2 "g O"_2) = "0.5000 mols O"_2$

No, it is not necessary to halve anything. You have from the masses the same mol/mol ratio as required in the reaction itself. You needed $"1 mol CH"_4(g)$ for every $"0.5 mol O"_2(g)$ , and that's what you have, basically.

You do have a bit extra $"O"_2$ than you need, though, so let's just be particular about it... The limiting reactant is $"CH"_4$ , since there is slightly less than twice the mols of $"CH"_4$ as $"O"_2$ , when the reaction requires exactly twice.

So:

$"0.9973 mols CH"_4 harr "0.4987 mols O"_2$

and $"0.0130 mols O"_2$ is in excess. That will remain, but we have to account for that in calculating the mols of the products by using $"CH"_4$ instead of $"O"_2$ to calculate the mols of products.

By the mol/mol ratio given in the reaction:

$"mols CH"_4 harr "0.9973 mols CO"(g)$

$"mols CH"_4 xx 2 harr "1.9947 mols H"_2(g)$

So, really the next thing we can calculate are the final and initial mols:

$n_(CO) + n_(H_2) = n_2$

$n_(CH_4) + n_(O_2) = n_1$

$=> "0.9973 mols CO" + "1.9947 mols H"_2 = n_2 = "2.9920 mols gas"$

$=> "0.9973 mols CH"_4 + "(0.4987 + 0.0130) mols O"_2 = n_1 = "1.5090 mols gas"$

That means the ratio of the mols can be related for a big picture using Gay-Lussac's Law:

$V_1/(n_1) = V_2/(n_2)$

$V_1/("1.5090 mols") = V_2/("2.9920 mols")$

By the molar volume, we can find each actual volume for the ideal gas combinations.

$V_1/"1.5090 mols" = "80.91 L"/"mol"$

$=> V_1 = "122.093 L"$

$V_2/"2.9920 mols" = "80.91 L"/"mol"$

$=> V_2 = "242.083 L"$

So, the change in volume is then:

$bb(DeltaV = V_2 - V_1)$

$= "242.083 L" - "122.093 L"$

$=$ $bb"119.99 L"$

To two sig figs, $color(blue)(DeltaV = "120 L")$ .

As a check, we can see whether the ideal gas law is still satisfied.

$PDeltaV stackrel(?" ")(=)DeltanRT$

$("1 bar")("120 L") stackrel(?" ")(=) ("2.9920 - 1.5090 mols gas")("0.083145 L"cdot"bar/mol"cdot"K")("973.15 K")$

$"120 L"cdot"bar" ~~ "119.993 L"cdot"bar"$

Yep, we're good.

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What is the change in volume for the reaction of $"16 g CH"_4$ and $"16 g O"_2$ at $700^@ "C"$ and $"1 bar"$ ? The reaction is $"CH"_4(g) + 1/2"O"_2(g) -> "CO"(g) + 2"H"_2(g)$ .

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What is the change in volume for the reaction of $"16 g CH"_4$ and $"16 g O"_2$ at $700^@ "C"$ and $"1 bar"$ ? The reaction is $"CH"_4(g) + 1/2"O"_2(g) -> "CO"(g) + 2"H"_2(g)$ .